Oracle Regex match all words ignoring order - regex

Question

I need find records matching all query phrases but ignoring their occurrence order.
For example, my query string is apple banana kiwi. Following values should be true.
I like apple, banana and kiwi
Banana, kiwi and apple are fruits
Following values should be false
He does not like kiwi
How to implement by SQL in Oracle 11?

Answer 1

In a modern regex engine you would use look-ahead assertions to combine the three conditions into one expression:
(?:.*?\bapple\b)(?:.*?\bbanana\b)(?:.*?\bkiwi\b)
Oracle does not support look-aheads, though, and that means you cannot write an expression that checks all three conditions at the same time(*).
Your options:
Split up the regular expression and combine multiple simple expressions with AND - this is the slowest variant, but it would work.
Dump regular expressions and use multiple LIKE clauses with AND - this will a little be faster than regex but but expression complexity is limited in comparison.
Set up a full text index on that table and use it - this will be the fastest variant, but expression complexity is limited compared to regex. It will be sufficient for a pure natural language keyword search, though, and it would support stemming and alternative word forms.
(*) Technically, academically, you can. You could write an expression that checks all possible permutations of your keywords, like this
A.*?B.*?C|B.*?C.*?A|C.*?A.*?B|...and so on|and so forth
Think whether you would call this an acceptable solution. Oh yeah and it would be slow as hell, too.

Answer 2

Here is an attempt:
with w as -- The words
(
select 'apple banana kiwi' words from dual
),
p as -- the patterns taken from the words
(
select regexp_substr(w.words, '\w+', 1, level) pattern
from w
connect by regexp_substr(w.words, '\w+', 1, level) is not null
),
r as -- the phrases to test
(
select 'I like apple, banana and kiwi' phrase from dual
union all
select 'Banana, kiwi and apple are fruits' phrase from dual
union all
select 'He does not like kiwi' phrase from dual
)
select r.phrase
case sum(case instr(upper(r.phrase), upper(p.pattern))
when 0 then 0
else 1 end)
when regexp_count(w.words, '\w+', 1) then 'true'
else 'false' end all_present
from r, p, w
group by r.phrase, w.words
;
And the result:
He does not like kiwi false
Banana, kiwi and apple are fruits true
I like apple, banana and kiwi true
The principle:
test for every pattern if it's in the phrase (by instr: if 0, it's not present, else it is)
group by phrase to sum this match
if this sum is equal to the number of words tested (here, 3), this is true